An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t) which we write hth, ttt, etc. For each outcome, let N be the random variable counting the number of heads in each outcome. For example, if the outcome is hht, then N (hht)=2. Suppose that the random variable X is defined in terms of N as follows: X=6N-2N²-4. The values of X are given in the table below. Outcome hth ttt tth tht hhh thh htt hht Value of X 0 -4 0 0 -4 0 0 0 Calculate the probabilities P(X=x) of the probability distribution of X. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row. Value x of X P(X=x) 010
An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t) which we write hth, ttt, etc. For each outcome, let N be the random variable counting the number of heads in each outcome. For example, if the outcome is hht, then N (hht)=2. Suppose that the random variable X is defined in terms of N as follows: X=6N-2N²-4. The values of X are given in the table below. Outcome hth ttt tth tht hhh thh htt hht Value of X 0 -4 0 0 -4 0 0 0 Calculate the probabilities P(X=x) of the probability distribution of X. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row. Value x of X P(X=x) 010
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter12: Probability
Section12.3: Conditional Probability; Independent Events; Bayes' Theorem
Problem 24E
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Transcribed Image Text:An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t) which we write hth, ttt, etc.
For each outcome, let N be the random variable counting the number of heads in each outcome. For example, if the outcome is hht, then N (hht) =2. Suppose
that the random variable X is defined in terms of N as follows: X=6N-2N²-4. The values of X are given in the table below.
Outcome hth ttt tth tht hhh thh htt hht
Value of X 0 -4 0 0 -4 0
Calculate the probabilities P(X=x) of the probability distribution of X. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in
the second row.
Value X of X
P(X=x)
0 0
00
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